【二叉树】【层次遍历】116. 填充每个节点的下一个右侧节点指针

原创已于 2025-06-18 15:47:34 修改 · 493 阅读

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#层次遍历

于 2024-01-04 02:38:04 首次发布

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该博客围绕二叉树层次遍历展开，具体针对题目“填充每个节点的下一个右侧节点指针”，给出了使用层次遍历的解法，并分别展示了Python和Java两种语言的实现。

题目

法1：层次遍历

Python

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
        if not root:
            return None
        queue = collections.deque()
        queue.append(root)
        while queue:
            tmp_size = len(queue)
            pre_node = None
            for i in range(tmp_size):
                cur_node = queue.popleft()
                if cur_node.left:
                    queue.append(cur_node.left)
                if cur_node.right:
                    queue.append(cur_node.right)
                if pre_node:
                    pre_node.next = cur_node
                pre_node = cur_node
        return root

Java

class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return root;
        }
        Queue<Node> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; ++i) {
                Node curNode = queue.poll();
                if (i < size - 1) {
                    curNode.next = queue.peek();
                } else {
                    curNode.next = null;
                }
                if (curNode.left != null) {
                    queue.offer(curNode.left);
                }
                if (curNode.right != null) {
                    queue.offer(curNode.right);
                }
            }
        }

        return root;
    }
}